Question

a path goes around a triangular park, as shown.
a. find the distance around the park to the nearest yard.
the distance is about yards.
b. a new path and a bridge are constructed from point q to the midpoint m of pr. find qm to the nearest yard.
qm ≈ yd
c. a man jogs from p to q to m to r to q and back to p at an average speed of 150 yards per minute. to the nearest tenth of a minute, about how long does it take him to travel the entire distance?
it takes about minutes

Explanation:

Step1: Find coordinates of points

Assume $Q=(0,0)$, $P=(0,50)$, $R=(80,0)$.

Step2: Calculate length of $PQ$

Using distance formula for two - points $(x_1,y_1)$ and $(x_2,y_2)$: $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $P(0,50)$ and $Q(0,0)$, $PQ=\sqrt{(0 - 0)^2+(50 - 0)^2}=50$ yards.

Step3: Calculate length of $QR$

For $Q(0,0)$ and $R(80,0)$, $QR=\sqrt{(80 - 0)^2+(0 - 0)^2}=80$ yards.

Step4: Calculate length of $PR$

For $P(0,50)$ and $R(80,0)$, $PR=\sqrt{(80 - 0)^2+(0 - 50)^2}=\sqrt{6400 + 2500}=\sqrt{8900}\approx94.3$ yards.

Step5: Find perimeter of the park (answer for part a)

The distance around the park (perimeter $P$) is $PQ+QR + PR=50 + 80+94.3 = 224.3\approx224$ yards.

Step6: Find mid - point $M$ of $PR$

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $P(0,50)$ and $R(80,0)$, $M=(\frac{0 + 80}{2},\frac{50+0}{2})=(40,25)$.

Step7: Calculate length of $QM$ (answer for part b)

For $Q(0,0)$ and $M(40,25)$, $QM=\sqrt{(40 - 0)^2+(25 - 0)^2}=\sqrt{1600 + 625}=\sqrt{2225}\approx47.2\approx47$ yards.

Step8: Calculate total distance of the jogging path

The jogging path is $PQ+QM+MR+RQ+QP$. $MR = PR/2\approx94.3/2 = 47.2$ yards. Total distance $D=50+47+47+80+50 = 274$ yards.

Step9: Calculate time taken for jogging (answer for part c)

Using the formula $t=\frac{d}{v}$, where $d$ is distance and $v$ is speed. Given $v = 150$ yards per minute, $t=\frac{274}{150}\approx1.8$ minutes.

Answer:

a. 224
b. 47
c. 1.8